 Solution

Differentiation of the Hubble parameter gives $\dot{H}+H^2=\ddot{a}/a$. The strong energy condition and the acceleration equation therefore implyAssume the present universe is expanding, $H_0>0$, and normalize the scale factor by $a(t_0)=1$. Since $\ddot{a}\le 0$, the past expanding branch has $\dot{a}(t)\ge \dot{a}(t_0)>0$. On this branch,Integrating from a past time $t$ to $t_0$ gives $H(t{)}^{-1}\le H_0^{-1}+t-t_0$. Thus, where the denominator is positive,The printed inequality needs this past-time qualification: for future times with $H>0$, the inequality reverses. Independently, a concave function lies below its tangent, soPositivity of $a$ prevents the past branch extending before the zero of this tangent. If $t_B$ is the beginning of the expanding universe, its age satisfies $\boxed{t_0-t_B\le H_0^{-1}}$. Without the normalization $a(t_0)=1$, the tangent bound instead reads $a(t)\le a(t_0)[1+H_0(t-t_0)]$.